MCQ Questions for Class 11 Commerce Applied Mathematics Limits And Continuity Quiz 5

The limit of

$x\sin { \left( { e }^{ \frac { 1 }{ x } } \right) }$ as

$x\rightarrow 0$

0%
is equal to $0$
0%
is equal to $1$
0%
is equal to $\cfrac { e }{ 2 }$
0%
does not exist

Explanation

$\displaystyle \lim _{ x\rightarrow 0 }{ x\sin { { e }^{ (1/x) } } }$
LHL

$=f(0-0)=\displaystyle \lim _{ x\rightarrow 0 }{ (-h)\sin { { e }^{ (-1/h) } } }$

$=-0\times \sin { \left( { e }^{ -\infty } \right) }$

$=-0\times \sin { (0)=0 }$

$=0\times$ (a finite number between

$-1$ to

$+1$ )

$\quad \left( \because -1\le \sin { x } \le 1 \right)$

$\because$ LHL

$=$ RHL

$\because$

$\displaystyle \lim _{ x\rightarrow 0 }{ x\sin { { e }^{ (1/x) } } }$ exist and equal to

$0$

The function represented by the following graph is.

0%
Differentiable but not continuous $x=1$
0%
Neither continuous nor Differentiable at $x=1$
0%
Continuous but not Differentiable at $x=1$
0%
Continuous but Differentiable at $x=1$

Explanation

Since there is no discontinuity in the graph so clearly function is continuous,

but is not differentiable at

$x=1$ , because graph of the function has a sharp point or kink.

Note: The point on the graph of any function where it has a sharp corner or kink, function is not differentiable .

If [x] denotes the greatest integer not exceeding

$x$ and if the function

$f$ defined by

$f(x)= \begin{cases}\dfrac{a+2\cos\,x}{x^2}&(x < 0) \\ b\,\tan \dfrac{\pi}{[x+4]}&(x \ge 0) \end{cases}$ is continuous at

$x=0$ , then the order pair (a, b) =

0%
$(-2, 1)$
0%
$(-2, -1)$
0%
$(-1, \sqrt{3})$
0%
$(-2,-\sqrt{3})$

Explanation

Since the function is continuous at

$x = 0, f(0^-) = f(0^+)$

$\Rightarrow \displaystyle \lim_ {x \rightarrow 0^-} \cfrac{a + 2\cos{x}}{x^2} = \displaystyle \lim_{x \rightarrow 0^+} b \ \tan \ \cfrac{\pi}{[x + 4]}$

For a finite limit,

$a$ has to be

$-2$ since the numerator has to be

$0$ for the limit to be of the form

$\dfrac{0}{0}$

Using L'Hospital rule, we get

$\displaystyle \lim_{x \rightarrow 0^-} \cfrac{-2 \ \sin{x}}{2x} = -1$

$\Rightarrow b \ \tan \ \cfrac{\pi}{4} = -1$

$\Rightarrow b = -1$

$\displaystyle \lim_{x\rightarrow 0}\dfrac {1 - \cos x}{x^{2}}$ is ____

0%
$2$
0%
$3$
0%
$\dfrac {1}{2}$
0%
$\dfrac {1}{3}$

Explanation

$\displaystyle \lim_{x\rightarrow 0}\dfrac {1 - \cos x}{x^{2}}=\lim_{x\to 0}\dfrac{2\sin^2\frac{x}{2}}{x^2}$

$=\displaystyle \lim_{x\rightarrow 0}\dfrac {2\sin^2\frac{x}{2}}{4\left(\frac{x}{2}\right)^{2}}=\dfrac{1}{2}\lim_{x\to 0}\left( \dfrac{\sin\frac{x}{2}}{\frac{x}{2}}\right)^2=\dfrac{1}{2}\cdot 1=\dfrac{1}{2}$ (using basic limit formula )

$\displaystyle \lim_{x\rightarrow \infty}\dfrac{x^3+1}{x^2+1}-(ax+b)=2$ , then

0%
$a=2$ and $b=-1$
0%
$a = 1$ and $b = 1$
0%
$a = 1$ and $b = -1$
0%
$a = 1$ and $b = -2$

Explanation

Given

$\displaystyle \lim _{ x\rightarrow \infty }{ \frac { { x }^{ 3 }+1 }{ { x }^{ 2 }+1 } -(ax+b)=2 }$

$\Rightarrow \displaystyle \lim _{ x\rightarrow \infty }{ (\frac { { (1-a)x }^{ 3 }-b{ x }^{ 2 }-ax-b+1 }{ { x }^{ 2 }+1 } )=2 }$

For the limit to exist, the coefficient of

$x^{3}$ must be zero because if it is not zero then the limit is infinite

$\Rightarrow a=1$

So the given one will reduce to

$\displaystyle \lim _{ x\rightarrow \infty }{ (\frac { -b{ x }^{ 2 }-ax-b+1 }{ { x }^{ 2 }+1 } )=2 }$

If we apply the limit , we get

$-b=2$

$\Rightarrow b=-2$

Therefore the correct option is

$D$

$\lim _ { x \rightarrow 0 } \dfrac { ( 27 + x ) ^ { 1 / 3 } - 3 } { 9 - ( 27 + x ) ^ { 2 / 3 } }$ equals :

0%
$- 1 / 6$
0%
$1 / 6$
0%
$1 / 3$
0%
$- 1 / 3$

Explanation

$\underset { x\rightarrow 0 }{ lt } \dfrac { { \left( 27+x \right) }^{ 1/3 }-3 }{ 9-{ \left( 27+x \right) }^{ 2/3 } }$

$=\underset { x\rightarrow 0 }{ lt } \dfrac { \dfrac { 1 }{ 3 } { \left( 27+x \right) }^{ -2/3 } }{ \dfrac { -2 }{ 3 } { \left( 27+x \right) }^{ -1/3 } }$

$=\underset { x\rightarrow 0 }{ lt } \dfrac { -1 }{ 2 } \dfrac { { \left( 27+x \right) }^{ 1/3 } }{ { \left( 27+x \right) }^{ 2/3 } }$

$=\dfrac { -1 }{ 2 } \left( \dfrac { 3 }{ 9 } \right) ,\quad \dfrac { -1 }{ 6 }$ [A]

The value of the constant

$\alpha$ and

$\beta$ such that

$\displaystyle \lim_{x\rightarrow \infty}\left(\displaystyle\frac{x^2+1}{x+1}-\alpha x-\beta\right)=0$ are respectively.

0%
$(1, 1)$
0%
$(-1, 1)$
0%
$(1,-1)$
0%
$(0, 1)$

Explanation

Simplifying the above expression gives us

$\displaystyle \lim_{x\rightarrow \infty}(\dfrac{x^{2}+1-\alpha x(x+1)-\beta(x+1)}{x+1})$

$=\displaystyle \lim_{x\rightarrow \infty}(\dfrac{x^{2}(1-\alpha)-x(\alpha +\beta)+1-\beta)}{x+1})$

$=0$ implies that

$Numerator<Denominator$ .
Hence coefficient of

$x^{2}$ will be 0. Therefore

$\alpha=1$ .
Hence

$\displaystyle \lim_{x\rightarrow \infty}(\dfrac{-x(1 +\beta)+1-\beta)}{x+1})=0$

Applying L'Hopital's Rule we get

$\displaystyle \lim_{x\rightarrow \infty}(\dfrac{-(1 +\beta)}{1})=0$
Or

$\beta+1=0$
Or

$\beta=-1$ .
Hence

$(\alpha,\beta)=1,-1$

What is

$\displaystyle \lim_{x \rightarrow 0 } x^2 \sin \left(\frac{1}{x}\right)$ equal to ?

0%
0
0%
1
0%
1/2
0%
Limit does not exist.

Explanation

$\displaystyle \lim _{ x\rightarrow 0 }{ { x }^{ 2 }\sin { \left( \cfrac { 1 }{ x } \right) } }$

$={ 0 }^{ 2 }\times \sin { \left( \cfrac { 1 }{ 0 } \right) } \quad \left[ -1\le \sin { \left( \cfrac { 1 }{ x } \right) } \le 1 \right]$

$=0$

If the function

$f(x)$ satisfies

$\displaystyle \lim_{x\rightarrow 1}\frac{f(x)-2}{x^2-1}=\pi$ , then

$\displaystyle \lim_{x\rightarrow 1}f(x)=$

0%
$2$
0%
$3$
0%
$1$
0%
$0$

Explanation

It is given that,

$\displaystyle \lim_{x\rightarrow 1}\frac{f(x)-2}{x^2-1}=\pi$ exists and equal to

$1$ .

Clearly the denominator is zero at

$x=1$ , so for above limit to exist

numerator should also be zero, as x approaches

$1$ .

Therefore,

$\displaystyle \lim_{x\to 1}f(x)=2$

Note: If

$\displaystyle \lim_{x\to 1}f(x) \neq 1$ , given limit will not exist

The limit of

$\left[\frac{1}{x^2}+\frac{(2013)^x}{e^x-1}-\frac{1}{e^x-1}\right]$ as

$x\rightarrow 0$

0%
Approaches $+\, \infty$
0%
Approaches $- \,\infty$
0%
Is equal to $log_e(2013)$
0%
Does not exist

Explanation

$\displaystyle \lim _{ x\rightarrow 0 }{ \left[ \cfrac { 1 }{ { x }^{ 2 } } +\cfrac { { \left( 2013 \right) }^{ x }-1 }{ { e }^{ x }-1 } \right] }$

$=\displaystyle \lim _{ x\rightarrow 0 }{ \left[ \cfrac { 1 }{ { x }^{ 2 } } +\cfrac { { \left( 2013 \right) }^{ x }-1 }{ x } \times \cfrac { x }{ { e }^{ x }-1 } \right] }$

$=\left( \displaystyle \lim _{ x\rightarrow 0 }{ \cfrac { 1 }{ { x }^{ 2 } } } \right) +\left( \log { 2013 } \right) \times 1\quad \quad \left[ \displaystyle \lim _{ x\rightarrow 0 }{ \cfrac { { a }^{ x }-1 }{ x } } =\log _{ e }{ a } ;\quad \displaystyle \lim _{ x\rightarrow 0 }{ \cfrac { { e }^{ x }-1 }{ x } } =1 \right]$

$=\infty +\log { 2013 } =\infty$

$x\rightarrow 0\Rightarrow$ Limit

$\rightarrow \infty$

$\displaystyle \lim_{x\rightarrow 0}\frac{log_e(1+x)}{3^x-1}=$ ____.

0%
$log_e3$
0%
$0$
0%
$log_3 e$
0%
$1$

Explanation

Given that:

$\displaystyle \lim_{x \rightarrow 0} \cfrac{\log_e(1 + x)}{3^x - 1}$

Using L'Hospital rule, we differentiate the numerator and denominator to find the limit.

We thus get

$\displaystyle \lim_{x \rightarrow 0} \cfrac{1}{(1 + x)(3^x)(\log_e3)}$

$= \cfrac{1}{1 \times 1 \times \log_e3}$

$= \cfrac{1}{\log_e3}$

$= \log_3e$

The limit of

$\displaystyle \sum_{n=1}^{1000}(-1)^nx^n$ as

$x\rightarrow \infty$

0%
does not exist
0%
exists and equals to 0
0%
exists and approaches $+\infty$
0%
exists and approaches $-\infty$

Explanation

$\sum _{ n=1 }^{ 1000 }{ { \left( -1 \right) }^{ n }{ x }^{ n } } =-x+{ x }^{ 2 }-{ x }^{ 3 }+{ x }^{ 4 }+....$

It is in GP

$S=\cfrac { -x\left( { (-x) }^{ 1000 }-1 \right) }{ (-x)-1 }$

$\displaystyle \lim _{ x\rightarrow \infty }{ \sum _{ n=1 }^{ 1000 }{ { \left( -1 \right) }^{ n }{ x }^{ n } } } =\displaystyle \lim _{ x\rightarrow \infty }{ S } =\displaystyle \lim _{ x\rightarrow \infty }{ \cfrac { x\left( { x }^{ 1000 }-1 \right) }{ x+1 } }$

$=\displaystyle \lim _{ x\rightarrow \infty }{ \cfrac { \left( { x }^{ 1000 }-1 \right) }{ 1+\cfrac { 1 }{ x } } } =\cfrac { { \infty }^{ 1000 }-1 }{ 1+0 } =\infty$

$as \left( x\rightarrow \infty ;\cfrac { 1 }{ x } \rightarrow \infty \right)$

Which one of the following statements is correct?

0%
$\displaystyle \lim_{x \rightarrow 0} (fog) (x)$ exists.
0%
$\displaystyle \lim_{x \rightarrow 0} (gof) (x)$ exists.
0%
$\displaystyle \lim_{x \rightarrow 0-} (fog) (x) = \displaystyle \lim_{x \rightarrow 0-} (gof) (x)$
0%
$\displaystyle \lim_{x \rightarrow 0+} (fog) (x) =\displaystyle \lim_{x \rightarrow 0-} (gof) (x)$

Explanation

$(fog) (x) = [\sin x]$

$\displaystyle \lim_{x\to 0^+} [\sin x]=0$ and

$\displaystyle \lim_{x\to 0^-} [\sin x]=-1$ . Hence,

$\displaystyle \lim_{x\to 0} fog(x)$ does not exist.

Now,

$gof(x)=\sin [x]$

$\displaystyle \lim_{x\to 0^+} \sin [x]=0$ and

$\displaystyle \lim_{x\to 0^-} \sin [x]=\sin (-1)$ . Hence,

$\displaystyle \lim_{x\to 0} gof(x)$ does not exist.

But,

$\displaystyle \lim_{x\to 0^+} fog(x)=\displaystyle \lim_{x\to 0^-} gof(x)$

The limit of

$\left\{\frac{1}{x}\sqrt{1-x}-\sqrt{1+\frac{1}{x^2}}\right\}$ as

$x\rightarrow 0$

0%
Does not exist
0%
Is equal to $\frac{1}{2}$
0%
Is equal to 0
0%
Is equal to 1

Explanation

Let

$p(x)=\cfrac { 1 }{ x } \sqrt { 1-x } ,q(x)=\sqrt { 1+\cfrac { 1 }{ { x }^{ 2 } } }$

$\lim _{ x\rightarrow { 0 }^{ + } }{ p\left( x \right) } =\lim _{ x\rightarrow { 0 }^{ + } }{ \cfrac { 1 }{ x } \sqrt { 1-x } } =\infty \times \sqrt { 1-10 }$

$=\infty$

$\lim _{ x\rightarrow { 0 }^{ - } }{ p\left( x \right) } =-\infty \times \sqrt { 1-0 } =-\infty$

$x\rightarrow { 0 }^{ + }\quad \cfrac { 1 }{ x } \rightarrow +\infty$

$x\rightarrow { 0 }^{ - }\quad \cfrac { 1 }{ x } \rightarrow -\infty$

$\Rightarrow p(x)$ is discontinuous as

$x\rightarrow 0$

Let

$f\left( x \right) =p\left( x \right) -q\left( x \right)$

Since

$p\left( x \right)$ is discontinuous

$f\left( x \right)$ is also discontinuous by Algebra of continuous functions.

$\displaystyle\lim_{x\rightarrow\frac{\pi}{6}}\frac{\sin\left(x-\displaystyle\frac{\pi}{6}\right)}{\sqrt{3-2cos x}}$ is equal to :

0%
$0$
0%
$\displaystyle\frac{1}{(\sqrt{3}-2)}$
0%
$1$
0%
$\infty$

Explanation

$\lim _{ x\rightarrow \frac { \pi }{ 6 } }{ \dfrac { \sin { (x-\cfrac { \pi }{ 6 } } ) }{ \sqrt { 3-2\cos { x } } } }$

$\cfrac { \pi }{ 6 }$ we get the numerator as

$\sin { (x-\cfrac { \pi }{ 6 } ) } = 0$

and the denominator as

$\sqrt { 3-2\cos { x } } =\sqrt { 3-\sqrt { 3 } }$

So,it is not an indeterminant form as we have

$\dfrac{0}{\sqrt{3 - \sqrt{3}} }$

Hence the answer will be zero.

Evaluate:

$\displaystyle\lim_{x\to 10}\dfrac{x^2-100}{x-10}$

0%
$10$
0%
$-5$
0%
$20$
0%
$5$

Explanation

$\displaystyle\lim_{x\to 10}\dfrac{x^2-100}{x-10}$

$\displaystyle\lim_{x\to 10}\dfrac{(x-10)(x+10)}{x-10}$

$\displaystyle\lim_{x\to 10}{x+10}$

$=10+10=20$

$\underset { x\rightarrow 0 }{ lim } \cfrac { { \left( 25 \right) }^{ x }-2\left( 15 \right)^ x+{ 9 }^{ x } }{ cos6x-cos2x }$ is equal to :

0%
$log \left ( \cfrac {5} {3} \right )$
0%
$\cfrac {1} {4} log 15$
0%
$- \cfrac {1} {16} \left( \cfrac {5} {3} \right) ^2$
0%
$log \left ( \cfrac {3} {5} \right )$

$\displaystyle \lim_{x\rightarrow \pi/4} \dfrac {\tan x - 1}{\cos 2x}$ is equal to

0%
$1$
0%
$0$
0%
$-2$
0%
$-1$

Explanation

$\displaystyle \lim_{x\rightarrow \pi/4} \dfrac {\tan x - 1}{\cos 2x} = \displaystyle \lim_{h\rightarrow 0}\dfrac {\tan \left (\dfrac {\pi}{4} + h\right ) - 1}{\cos 2\left (\dfrac {\pi}{4} + h\right )}$

$\left [\because x = \dfrac {\pi}{4} + h\right ]$

$= \displaystyle \lim_{h\rightarrow 0} \dfrac {\left (\dfrac {1 + \tan h}{1 - \tan h}\right ) - 1}{\cos \left (\dfrac {\pi}{2} + 2h\right )}$

$= \displaystyle \lim_{h\rightarrow 0} \dfrac {1 + \tan h - 1 + \tan h}{-\sin 2h (1 - \tan h)}$

$= \displaystyle \lim_{h\rightarrow 0} \dfrac {-2\tan h}{2\sin h \cos h (1 - \tan h)}$

$= \displaystyle \lim_{h\rightarrow 0} \dfrac {-1}{\cos^{2} h (1 - \tan h)} = -1$

Hence, option D is correct.

$\underset{x\to 0}{\lim}\dfrac{x^a\sin^b x}{\sin(x^c)}, a, b, c, \in R \sim \{0\}$ exists and has non-zero value, then

0%
$a,b,c$ are in A.P
0%
$a,b,c$ are in G.P
0%
$a,b,c$ are in H.P
0%
none of these

Explanation

Given, limit is

$\lim_{x\rightarrow 0}\dfrac {x^a\sin^bx}{\sin (x^c)}$

$\underset{x\to 0}{\lim} x^a.\dfrac{\left(\dfrac{\sin x}{x}\right)^b}{\left(\dfrac{\sin(x^c)}{x^c}\right)}.\dfrac{x^b}{x^c}$

$=\underset{x\to 0}{\lim} x^{a+b-c} \dfrac{\left(\dfrac{\sin x}{x}\right)^b}{\left(\dfrac{\sin(x^c)}{x^c}\right)}$

The above limit non-zero valu only when

$a+b-c = 0$

$\displaystyle \lim_{x\rightarrow 3} = \dfrac {\sqrt {x} -\sqrt {3}}{\sqrt {x^{2} - 9}}$ is equal to

0%
$1$
0%
$3$
0%
$\sqrt {3}$
0%
$-\sqrt {3}$
0%
$0$

Explanation

The value of

$\displaystyle \lim_{x\rightarrow 3} = \dfrac {\sqrt {x} -\sqrt {3}}{\sqrt {x^{2} - 9}}$ is

$= \displaystyle \lim_{x\rightarrow 3} \dfrac {\sqrt {x} - \sqrt {3}}{\sqrt {x - 3}\sqrt {x} + 3} \times \dfrac {\sqrt {x} + \sqrt {3}}{\sqrt {x} + \sqrt {3}}$

$= \displaystyle \lim_{x\rightarrow 3} \dfrac {x - 3}{\sqrt {x - 3} \sqrt {x + 3}}\times \dfrac {1}{\sqrt {x} + \sqrt {3}}$

$= \displaystyle \lim_{x\rightarrow 3} \dfrac {\sqrt {x - 3}}{\sqrt {x + 3}} \times \dfrac {1}{\sqrt {x} + \sqrt {3}} = 0$

The value of

$\displaystyle \lim _{ x\rightarrow \pi /6 }{ \cfrac { 2\sin ^{ 2 }{ x } +\sin { x } -1 }{ 2\sin ^{ 2 }{ x } -3\sin { x } -1 } }$ is

0%
$3$
0%
$-3$
0%
$6$
0%
$0$

Explanation

$L=\lim _{ x\rightarrow { \pi }/{ 6 } }{ \dfrac { 2\sin ^{ 2 }{ x } +\sin { x } -1 }{ 2\sin ^{ 2 }{ x } -3\sin { x } -1 } } \\ L=\dfrac { 2\sin ^{ 2 }{ \left( \dfrac { \pi }{ 6 } \right) } +\sin { \left( \dfrac { \pi }{ 6 } \right) } -1 }{ 2\sin ^{ 2 }{ \left( \dfrac { \pi }{ 6 } \right) } -3\sin { \left( \dfrac { \pi }{ 6 } \right) } -1 } \\ L=\dfrac { 2.\dfrac { 1 }{ 4 } +\dfrac { 1 }{ 2 } -1 }{ 2.\dfrac { 1 }{ 4 } -3.\dfrac { 1 }{ 2 } -1 } =\dfrac { 0 }{ -2 }= 0$

So option $D$ is correct

$f\left( x \right) =\begin{vmatrix} \sin { x } & \cos { x } & \tan { x } \\ { x }^{ 3 } & { x }^{ 2 } & x \\ 2x & 1 & 1 \end{vmatrix}$ , then

$\displaystyle\lim _{ x\rightarrow 0 }{ \dfrac { f\left( x \right) }{ { x }^{ 2 } } }$ is

0%
$-1$
0%
$3$
0%
$1$
0%
Zero

Explanation

$f(x)=\begin{vmatrix} sinx & cosx & tanx \\ { x }^{ 3 } & { x }^{ 2 } & x \\ 2x & 1 & 1 \end{vmatrix}\\ \Longrightarrow f(x)=sinx({ x }^{ 2 }-x)-cosx({ x }^{ 3 }-2{ x }^{ 2 })+tanx({ x }^{ 3 }-2{ x }^{ 3 })\\ \Longrightarrow f(x)={ x }^{ 2 }sinx-xsinx-{ x }^{ 3 }cosx+2{ x }^{ 2 }cosx-{ x }^{ 3 }tanx\\ \Longrightarrow \dfrac { f(x) }{ { x }^{ 2 } } =sinx-xcosx+2cosx-xtanx-\dfrac { sinx }{ x } \\ \Longrightarrow \underset { x\longrightarrow 0 }{ lim } \dfrac { f(x) }{ { x }^{ 2 } } =\underset { x\longrightarrow 0 }{ lim } sinx-\underset { x\longrightarrow 0 }{ lim } xcosx+\underset { x\longrightarrow 0 }{ lim } 2cosx-\underset { x\longrightarrow 0 }{ lim } xtanx-\underset { x\longrightarrow 0 }{ lim } \dfrac { sinx }{ x } \\ \Longrightarrow \underset { x\longrightarrow 0 }{ lim } \dfrac { f(x) }{ { x }^{ 2 } } =0-0+2-0-1\\ \Longrightarrow \underset { x\longrightarrow 0 }{ lim } \dfrac { f(x) }{ { x }^{ 2 } } =1\\ \\$

$\displaystyle \lim _{ n\rightarrow \infty }{ \cfrac { n.{ 3 }^{ n } }{ n{ \left( x-2 \right) }^{ n }+n.{ 3 }^{ n+1 }-{ 3 }^{ n } } } =\cfrac { 1 }{ 3 }$ , then the range of

$x$ is (When

$n\in N$ )

0%
$[2,\ 5)$
0%
$\left( 1,\ 5 \right)$
0%
$\left( -1,\ 5 \right)$
0%
$\left( -\infty ,\ \infty \right)$

Explanation

$\displaystyle\lim_{n\rightarrow \infty}\dfrac{n.3^n}{n(x-2)^n+n3^{n+1}-3^n}=\displaystyle\lim_{n\rightarrow \infty}\dfrac{1}{\left(\dfrac{x-2}{3}\right)^n+3-\dfrac{1}{n}}=\dfrac{1}{3}$

We know

$\displaystyle\lim_{n\rightarrow \infty}\dfrac{1}{n}=0$

then we want

$\dfrac{|x-2|}{3} < 1$

$\rightarrow \because$ if

$a < 1, a^n=0$ for

$n\rightarrow \infty$

$x < 5$

$x > -1$

$\therefore (-1, 5)$ .

1160898_708152_ans_ef33ab65ae014854a1119d49d198e2b4.jpg

$\lim _{ x\rightarrow 3 }{ \left( { x }^{ 3 }-4 \right) /\left( x+1 \right) } =$

0%
$(4/23)$
0%
$(2/23)$
0%
$(1/8)$
0%
$(23/4)$

$\displaystyle\lim_{x\rightarrow \frac{\pi}{2}}(\pi - 2x^{\cos x})$ is equal to :

0%
$\pi+2$
0%
$\pi-2$
0%
e
0%
$e^{-1}$

Explanation

$\displaystyle\lim_{x\rightarrow \frac{\pi}{2}}(\pi - 2x^{\cos x})$

$=\pi - 2\left(\dfrac{\pi}{2}\right)^{\cos \frac{\pi}{2}}$

$=\pi-2\left(\dfrac{\pi}{2}\right)^{0}$

$=\pi-2$

Determine the value of k for which the following function is continuous at

$x=3$ .

$f(x)=\dfrac{x^2-9}{x-3}, x \neq 3$

$f(x)=k, x=3$

0%
2
0%
4
0%
6
0%
8

Explanation

Since f(x) is continuous at

$x=3$ .

Therefore,

$\displaystyle\lim_{x\rightarrow 3} f(x) =f(3)$

$\displaystyle\lim_{x\rightarrow 3} f(x)=k$

$\displaystyle\lim_{x\rightarrow 3} \dfrac{x^2-9}{x-3}=k$

$\displaystyle\lim_{x\rightarrow 3} \dfrac{(x+3)(x-3)}{x-3}=k$

$\displaystyle\lim_{x\rightarrow 3} (x+3)=k$

$k=6$

$f(x) = \begin{vmatrix} \cos x& x & 1\\ 2\sin x & x^{2} & 2x\ \\ \tan x & x & 1\end{vmatrix}$ , then

$\displaystyle \lim_{x\rightarrow 0} \dfrac {f'(x)}{x}$ .

0%
Exists and is equal to $-2$
0%
Does not exist
0%
Exist and is equal to $0$
0%
Exists and is equal to $2$

Explanation

$f(x) = \begin{vmatrix} \cos x& x & 1\\ 2\sin x & x^{2} & 2x\ \\ \tan x & x & 1\end{vmatrix}$

$=\cos x(x^{2}-2x^{2})-x(2\sin x-2x\tan x)+1(2x\sin x-x^{2}\tan x)$

$=-x^{2}\cos x-2x\sin x+2x^{2}\tan x+2x\sin x-x^{2}\tan x$

$=x^{2}\tan x-x^{2}\cos x$

$=x^{2}(\tan x-\cos x)$

$f'(x)=2x(\tan x-\cos x)+x^{2}(\sec^{2}x+\sin x)$

$\therefore \displaystyle \lim_{x\rightarrow 0} \dfrac{f'(x)}{x}=\lim_{x\rightarrow 0}\dfrac{2x(\tan x-\cos x)+x^{2}(\sec^2{x}+\sin x)}{x}$

$=\displaystyle \lim_{x\rightarrow 0} 2(\tan x-\cos x)+x(\sec^{2}x+\sin x)$

$=2(0-1)+0=-2$

Hence,

$\therefore \displaystyle \lim_{x\rightarrow 0} \dfrac{f'(x)}{x}=-2$

Which of the following statements are true for the function

$f(x)$ defined for

$-1\leq x \leq 3$ in the figure shown?

0%
$\lim_\limits{x\to-1^+} f(x) = 1$
0%
$\lim_\limits{x\to2} f(x)$ does not exist
0%
$\lim_\limits{x\to1^-} f(x) = 2$
0%
$\lim_\limits{x\to 0^+} f(x) =\lim_\limits{x\to 0^-} f(x)$

Explanation

From the picture it is clear that when

$x\to 1^+$ ,

$f(x)=1$ i.e., we are getting a neighbour-hood

$1<x\leq 3$ where

$f(x)=1$ .

So,

$\lim\limits_{x\to 1^+}f(x)=1$ .

So, option (A) is correct.

Also as

$x\to 1^{-}$ ,

$f(x)=2$ , so, option (C) is also correct.

$2$ where

$f(x)=1$ and

$f(x)=2$ as well, so

$\lim\limits_{x\to 2}f(x)$ does not exist.

Again (D) is also correct as we have a small neighbour-hood of

$0$ where the limits of

$f(x)$ exists.

$\dfrac{\displaystyle \lim_{h \rightarrow 0}(h+1)^2}{\displaystyle \lim_{h\rightarrow 0}(1+h)^{2/h}}$ is equal to

0%
$e^1$
0%
$e^{-2}$
0%
$e^2$
0%
$e^{-1}$

$\mathop {\lim }\limits_{x \to 0} {{(1 - \cos 2x)(3 + \cos x)} \over {x\tan 4x}}$ is equal to

0%
2
0%
1/2
0%
4
0%
3

Explanation

$\lim _{ x\rightarrow 0 }{ \cfrac { (1-\cos { 2x } )(3+\cos { x } ) }{ x\tan { 4x } } } \\ \Rightarrow \cfrac { (1-1+2\sin { ^{ 2 }{ x } } )(3+\cos { x } ) }{ x\times 4x } \times (\cfrac { 4x }{ \tan { 4x } } )\\ \Rightarrow \cfrac { 2\sin { ^{ 2 }{ x } } }{ 4{ x }^{ 2 } } \times (3+\cos { x } )\times 1\\ \Rightarrow \cfrac { 1 }{ 2 } { (\cfrac { \sin { x } }{ x } ) }^{ 2 }\times (3+\cos { x } )\times 1\\ \Rightarrow \cfrac { 1 }{ 2 } \times 4\times 1=2$

Evaluate the limit:

$\displaystyle \lim_{x \to 0} \left( \frac{x- \sin x}{x}\right) \sin \left(\frac{1}{x} \right)$

0%
$1$
0%
$-1$
0%
Does not exist
0%
$0$

Explanation

$\lim _{ x\rightarrow 0 }{ (\cfrac { x-\sin { x } }{ x } ) } \sin { (\cfrac { 1 }{ x } ) } \\ \lim _{ x\rightarrow 0 }{ (\cfrac { x-\sin { x } }{ x } ) } \times \lim _{ x\rightarrow 0 }{ \sin { (\cfrac { 1 }{ x } ) } } \\ (1-\cos { x } )\times (Value\quad between\quad -1\quad and\quad 1)\\ 0\times Value\quad between\quad (-1\& 1)\\ 0$

$\lim _{ x\rightarrow { 0 }^{ + } }{ \left( { \left( x\cos { x } \right) }^{ x }+{ \left( \cos { x } \right) }^{ \frac { 1 }{ \ln { x } } }+{ \left( x\sin { x } \right) }^{ x } \right) }$ is equal to

0%
$2$
0%
$2+e$
0%
$2+\dfrac { 1 }{ e }$
0%
$3$

Explanation

$\lim_{x\rightarrow 0^+}(x\cos x)^x$

$\lim_{x\rightarrow 0^+} x^x.\cos ^x x$

$\lim_{x\rightarrow 0^+} e^{x\ln x}.\cos ^x x$

$1\times 1$

$1$

$\therefore \lim_{x\rightarrow 0^+}(x\cos x)^x=1$

------------------------------------------------------

$\lim_{x\rightarrow 0^+}(x\sin x)^x$

$\lim_{x\rightarrow 0^+} x^x.\sin ^x x$

$\lim_{x\rightarrow 0^+} e^{x\ln x}.\sin ^x x$

$1\times 1$

$1$

$\therefore \lim_{x\rightarrow 0^+}(x\sin x)^x=1$

---------------------------------------------------------

$\lim_{x\rightarrow 0^+} (\csc x)^{\frac{1}{\ln x}}$

$\lim_{x\rightarrow 0^+} (\dfrac{1}{\sin x})^{\frac{1}{\ln x}}$

$y= \lim_{x\rightarrow 0^+} (\dfrac{1}{\sin x})^{\frac{1}{\ln x}}$

$\ln y =\lim_{x\rightarrow 0^+} \dfrac{1}{\ln x}(-\ln (\sin x))$

$\ln y =\lim_{x\rightarrow 0^+} \dfrac{1}{\frac{1}{x}}\left ( -\frac{\cos x}{\sin x} \right )$

$\ln y =\lim_{x\rightarrow 0^+}(-\cos x)$

$\ln y = -1$

$y=e^{-1}$

$y=\dfrac{1}{e}$

$\therefore \lim_{x\rightarrow 0^+} (\csc x)^{\frac{1}{\ln x}}=\dfrac{1}{e}$

------------------------------------------------------------------------------------

$\lim_{x\rightarrow 0^+} ((x\cos x)^x+(\csc x)^{\frac{1}{\ln x}}+(x\sin x)^x)$

$=1+\dfrac{1}{e}+1$

$=2+\dfrac{1}{e}$

If the function

$f(x)=\dfrac{e^{x^{2}}-\cos x}{x^{2}}$ for

$x \neq 0$ continuous at

$x=0$ then

$f(0)=$

0%
$\dfrac{1}{2}$
0%
$\dfrac{3}{2}$
0%
$2$
0%
$\dfrac{1}{3}$

Explanation

Here , function is continuous ,

So, left limit and right limit boh are equal and equals to the value of

$f(x)$ at

$x=0$

Therefore

$\displaystyle \lim_{x\rightarrow0} f(x)=\lim_{x\rightarrow0} \dfrac{e^{x^2}-cosx}{x^2}$

Solving limit by derivative approach,

=>

$\displaystyle \lim_{x\rightarrow0} f(x)=\lim_{x\rightarrow0} \dfrac{e^{x^2}.2x+sinx}{2x}$

=>

$\displaystyle \lim_{x\rightarrow0} f(x)=\lim_{x\rightarrow0} \dfrac{2e^{x^2}+e^{x^2}.x^2+cosx}{2x}$

Putting

$x=0$ , we get

=>

$f(0)=\dfrac{2+0+1}{2}=\dfrac{3}{2}$

The function $f : R /{0} \rightarrow R$ given by $f(x) = \dfrac{1}{x} - \dfrac{2}{e^{2x} -1}$ can be made continuous at $x=0$ by
defining $f(0)$ as

0%
0
0%
1
0%
2
0%
-1

Explanation

Given

$f(x)=\dfrac{1}{x}-\dfrac{2}{e^{2x}-1}$

For

$f(x)$ to be continuous at

$x=0$ , its limit should exists at

$x=0$ and must be finite.

Hence

$f(0)=\lim\limits_{x\to0}\dfrac{e^{2x}-1-2x}{x(e^{2x}-1)}$

$=\lim\limits_{x\to0}\dfrac{1+2x+\frac{4x^2}{2!}+\frac{8x^3}{3!}.....-1-2x}{x+2x^2+\frac{4x^3}{2!}.....-x}=\dfrac{\frac{4x^2}{2!}}{2x^2}=1$ ......[neglecting higher power of

$x$ ]

$\lim\limits_{x\to 0}\dfrac{e^{\sin x}-1}{x}=$

0%
$0$
0%
$1$
0%
$-1$
0%
none of these

Explanation

Now,

$\lim\limits_{x\to 0}\dfrac{e^{\sin x}-1}{x}$

$=\lim\limits_{x\to 0}\dfrac{\dfrac{e^{\sin x}-1}{\sin x}}{\dfrac{x}{\sin x}}$

$=\dfrac{\lim\limits_{\sin x\to 0}\dfrac{e^{\sin x}-1}{\sin x}}{\dfrac{1}{\lim\limits_{x\to 0}\dfrac{\sin x}{ x}}}$

$=\dfrac{1}{1}$

$=1$ .

Let

${P_n} = \prod\limits_{k = 2}^n {\left( {1 - {1 \over {{}^{{}^{k + 1}}{C_2}}}} \right)} .$ If

$\mathop {\lim }\limits_{x \to \infty } {P_n}$ can be expressed as lowest rational in the form

$\dfrac { a }{ b }$ , then value of

$(a+b)$ is __________.

0%
12
0%
4
0%
8
0%
10

$\mathop {\lim }\limits_{x \to 0} {\left( {\cos x + a\sin bx} \right)^{\frac{1}{x}}} = {e^2}$ then
the possible values of

$'a'\& 'b'are:$

0%
$a = 1,b = 2$
0%
$a = 2,b = 1$
0%
$a = 3,b = 2/3$
0%
$a = 2/3,b = 3$

Explanation

$\mathop {\lim }\limits_{x \to 0} {\left( {\cos x + a\sin bx} \right)^{\frac{1}{x}}}$ is of form

$1^{\infty}$ , so its limit will be

$e^k$ , where

$k=\lim\limits_{x\to0}\dfrac{1}{x}(cosx+asinbx-1)=\lim\limits_{x\to0}\dfrac{-sinx+abcosbx}{1}=ab=2$

Hence all possible combination of

$a$ and

$b$ are possible whose product is

$2$

$\mathop {\lim}\limits_{x \to \frac{\pi}{2}} \tan x =$

0%
$\infty$
0%
$- \infty$
0%
$0$
0%
does not exist

Explanation

$L=\displaystyle\lim_{x\rightarrow \frac{\pi}{2}}\tan x$

$\Rightarrow$

$L=\tan\dfrac{\pi}{2}$ [ Applying limit ]

We know that value of

$\tan\dfrac{\pi}{2}$ is does not exist.

$\therefore$

$\displaystyle\lim_{x\rightarrow \frac{\pi}{2}}\tan x=$ does not exist

Suppose the function

$f(x)-f(2x)$ has the derivative

$5$ at

$x=1$ and derivative

$7$ at

$x=2$ . The derivative of the function

$f(x)-f(4x)$ at

$x=1$ , has the value equal to?

0%
$19$
0%
$9$
0%
$17$
0%
$14$

$\lim _{ x\rightarrow 0 }{ \log _{ \left( \tan ^{ 2 }{ x } \right) }{ \left( \tan ^{ 2 }{ 2x } \right) = } }$

0%
$1$
0%
$2$
0%
$\dfrac {1}{2}$
0%
$Does\ not\ exist$

Explanation

$\mathop {\lim }\limits_{x \to 0} {\log _{{{\tan }^2}x}}\left( {{{\tan }^2}2x} \right)$

Let,

$\begin{array}{l} y={ \log _{ { { \tan }^{ 2 } }x } }{ \tan ^{ 2 } }2x \\ =\log { \tan ^{ 2 } } 2x={ \left( { { { \tan }^{ 2 } }x } \right) ^{ y } } \\ =2\log \tan 2x=2y\log \tan x \\ \Rightarrow y=\dfrac { { \log \tan 2x } }{ { \log \tan x } } \\ \Rightarrow { \lim }_{ x\to 0 }y={ \lim }_{ x\to 0 }\dfrac { { \log \tan 2x } }{ { \log \tan x } } \end{array}$

Using L-Hospital Rule

$\begin{array}{l} \Rightarrow { \lim }_{ x\to 0 }y={ \lim }_{ x\to 0 }\dfrac { { \dfrac { 1 }{ { \tan 2x } } \times { { \sec }^{ 2 } }2x\times 2 } }{ { \dfrac { 1 }{ { \tan x } } { { \sec }^{ 2 } }x } } \\ \Rightarrow { \lim }_{ x\to 0 }y={ \lim }_{ x\to 0 }\dfrac { { 2{ { \sec }^{ 2 } }2x } }{ { \tan 2x } } \times \dfrac { { \tan x } }{ { { { \sec }^{ 2 } }x } } \\ \Rightarrow { \lim }_{ x\to 0 }y={ \lim }_{ x\to 0 }\dfrac { { 2\tan x } }{ { \tan 2x } } \, \, \, \, \, \, \left( { { \lim }_{ x\to 0 }\sec x=4 } \right) \end{array}$

$\begin{array}{l} \Rightarrow { \lim }_{ x\to 0 }y={ \lim }_{ x\to 0 }\dfrac { { 2\dfrac { { \tan x } }{ x } \times x } }{ { \dfrac { { \tan 2x } }{ { 2x } } \times 2x } } \\ \Rightarrow { \lim }_{ x\to 0 }y={ \lim }_{ x\to 0 }\, \, \, 1\, \, \left( { { \lim }_{ x\to 0 }\dfrac { { \tan x } }{ x } =1 } \right) \\ \Rightarrow { \lim }_{ x\to 0 }{ \log _{ { { \tan }^{ 2 } }x } }\left( { { { \tan }^{ 2 } }2x } \right) =1 \end{array}$

The value of

$\displaystyle \lim_{x\rightarrow 0}\dfrac {\sqrt {1-\cos 2x}}{x}$ equals

0%
$0$
0%
$1$
0%
$\sqrt {2}$
0%
$Does\ not\ exist$

$l = \lim_{x \rightarrow 0} \dfrac{ x(1+ a\ cos x) - b\ sinx}{x^3}$ is finite,

where

$l \in R$ , then

0%
(a) $(a,b) = (-1,0)$
0%
(b) a & b are any real numbers
0%
(c) $l = 0$
0%
(d) $l = \frac{1}{2}$

Explanation

For given limit to exist, it should be of form

$\dfrac{0}{0}$ .

Now for limit, using L'Hopital rules (differentiating

$3$ time because denominator has power

$3$ ):

$l=\lim\limits_{x\to0}=\dfrac{x(1+acosx)-bsinx}{x^3}=\lim\limits_{x\to0}\dfrac{1-axsinx+acosx-bcosx}{3x^2}$

For limit to be finite:

$1+(a-b)=0\implies a-b=-1$ ........[1]

Again differentiating:

$l=\lim\limits_{x\to0}\dfrac{-axcosx-asinx-asinx+bsinx}{6x}$

Again differentiating:

$l=\lim\limits_{x\to0}\dfrac{-acosx+axsinx-acosx-acosx+bcosx}{6}=\dfrac{-a-a-a+b}{6}$

Now for

$b=0$ ;

$a=-1+b=-1$

So,

$l=\dfrac{-3a}{6}=\dfrac{1}{2}$

$\underset{h \rightarrow 0}{lim} \dfrac{\sqrt{x + h} -\sqrt{x}}{h}$ is equal to

0%
$\sqrt{x}$
0%
$\dfrac{1}{2 \sqrt{x}}$
0%
$2 \sqrt{x}$
0%
$\dfrac{1}{\sqrt{x}}$

Explanation

The given limit can be written as,

$\underset { h\rightarrow 0 }{ lim } \dfrac { \sqrt { x+h } -\sqrt { x } }{ h } =\underset { x+h\rightarrow x }{ lim } \dfrac { \sqrt { x+h } -\sqrt { x } }{ (x+h)-x }$

We know that,

$\underset { x\rightarrow a }{ lim } \dfrac { { x }^{ n }-{ a }^{ n } }{ x-a } =n{ a }^{ n-1 }$

So, the given limit is,

$=\dfrac { 1 }{ 2 } { x }^{ \left( \dfrac { 1 }{ 2 } -1 \right) }=\dfrac { 1 }{ 2\sqrt { x } }$

Evaluate:

$\lim\limits_{x\to 0}(1+ax)^{\dfrac{1}{x}}$

0%
$e^a$
0%
$e^{\dfrac{1}{a}}$
0%
$1$
0%
None of these

Explanation

Now,

$\lim\limits_{x\to 0}(1+ax)^{\dfrac{1}{x}}$

$=\left(\lim\limits_{ax\to 0}(1+ax)^{\dfrac{1}{ax}}\right)^a$

$=e^a$ .

The value of

$\lim\limits_{x\to 0}\dfrac{1-\cos x}{x^2}$

0%
$\dfrac{1}{2}$
0%
$\dfrac{1}{4}$
0%
$2$
0%
none of these

Explanation

Now,

$\lim\limits_{x\to 0}\dfrac{1-\cos x}{x^2}$

$=\lim\limits_{x\to 0}\dfrac{2\sin^2\dfrac{x}{2}}{x^2}$

$=\dfrac{1}{2}\times\left(\lim\limits_{\dfrac{x}{2}\to 0}\dfrac{\sin\dfrac{x}{2}}{\dfrac{x}{2}}\right)^2$

$=\dfrac{1}{2}\times 1$

$=\dfrac{1}{2}$ .

$\displaystyle\mathop {\lim }\limits_{\ x \to 0} \cos \frac{x}{2}\cos \frac{x}{{{2^2}}}\cos \frac{x}{{{2^3}}}......\cos \frac{x}{{{2^n}}}$ is equal to

0%
$1$
0%
$-1$
0%
$\dfrac{{\sin x}}{x}$
0%
$\dfrac{x}{{\sin \,x}}$

Explanation

$\displaystyle\lim_{x\rightarrow \infty}\left(\cos\dfrac{x}{2}\cos \dfrac{x}{2^{2}}, \cos\dfrac{x}{2^{3}}....... \cos\dfrac{x}{2^{x}}\right)$

$\sin 2\theta =2\sin \theta\cos\theta$

$2\theta=x$

$=\sin x=2\left(\sin\dfrac{x}{2}\right)\cos\dfrac{x}{2}$

$\theta=\dfrac{x}{2}$

$=2\left(2\sin \dfrac{x}{4}\cos \dfrac{x}{4}\right)\cos \dfrac{x}{2}$

$2\theta=\dfrac{x}{2}$

$\theta=\dfrac{x}{4}$

$=8\left(\left(\sin\dfrac{x}{8}\right)\cos\dfrac{x}{8}\cos\dfrac{x}{4}\cos\dfrac{x}{2}\right)$

$\dfrac{x}{4}=2\theta$

$\theta=\dfrac{x}{8}$

$=16\left(\sin \dfrac{x}{16}\cos\dfrac{x}{4}\cos\dfrac{x}{8}\cos\dfrac{x}{4}\cos\dfrac{x}{2}\right)\theta=\dfrac{x}{16}$

$=16\left(\sin\dfrac{x}{2^{4}}\cos\dfrac{x}{2^{4}}\cos\dfrac{x}{2^{3}}\cos \dfrac{x}{2^{2}}\cos\dfrac{x}{2}\right)$

$\sin x=\left(2^{n}\sin\dfrac{x}{2^{x}}\right)\cos\dfrac{x}{2x}\cos\dfrac{x}{2^{x-1}}.......\cos\dfrac{x}{2^{3}}\cos\dfrac{x}{2^{2}}\cos\dfrac{x}{2}$

$\left.\dfrac{\sin x}{2^{x}\sin\dfrac{x}{2^{x}}}\right]=\cos \dfrac{x}{2}\cos\dfrac{x}{2^{2}}\cos\dfrac{x}{2^{3}}..... \cos\dfrac{x}{2^{x}}$

$\displaystyle\lim_{x\rightarrow \infty}\dfrac{\sin x}{2^{n}\sin\left(\dfrac{x}{2^{x}}\right)}\lim_{x\rightarrow 0}\dfrac{\sin x}{x}=1$

$\overline { \infty } =0$

$\displaystyle\lim_{x\rightarrow \infty}\dfrac{\sin x}{2^{x}\dfrac{\sin\left(\dfrac{x}{2^{x}}\right)}{\left(\dfrac{x}{2^{x}}\right)}\times \left(\dfrac{x}{2^{x}}\right)}=\left.\dfrac{\sin x}{x}\right]\lim_{x\rightarrow \infty}\dfrac{\dfrac{1}{\sin\left(\dfrac{x}{2^{x}}\right)}}{\dfrac{x}{2^{x}}}$

$=\dfrac{\sin x}{x}$

$\lim\limits_{x\to 0}\dfrac{1-\cos x }{x^2}=$

0%
$4$
0%
$2$
0%
$\dfrac{1}{2}$
0%
$1$

Explanation

Now,

$\lim\limits_{x\to 0}\dfrac{1-\cos x}{x^2}$

$=\lim\limits_{x\to 0}\dfrac{2\sin^2\dfrac{x}{2}}{x^2}$

$=\dfrac{1}{2}\lim\limits_{x\to 0}\dfrac{\sin^2\dfrac{x}{2}}{\dfrac{x^2}{4}}$

$=\dfrac{1}{2}\left(\lim\limits_{\dfrac{x}{2}\to 0}\dfrac{\sin\dfrac{x}{2}}{\dfrac{x}{2}}\right)^2$

$=\dfrac{1}{2}.1$

$=\dfrac{1}{2}$ .

$\lim _{ x\rightarrow 1 }{ \dfrac { \sqrt { 1-\cos { 2\left( x-1 \right) } } }{ x-1 } }$

0%
Exists and it equals $\sqrt {2}$
0%
Exists and it equals $-\sqrt {2}$
0%
Does not exist because $x-1\rightarrow 0$
0%
Does not exist because left hand limit is not equal to right hand limit

$\underset{x \to 0}{\lim}\dfrac{\sin [ \cos x]}{1+[\cos x]}$ is

0%
$1$
0%
$0$
0%
does not exist
0%
$2$

$\mathop {\lim }\limits_{x \to 0} \dfrac{{{e^{4x}} - 1}}{x}$

0%
$1$
0%
$3$
0%
$4$
0%
$2$

Explanation

We have,

$\underset{x\to 0}{\mathop{\lim }}\,\left( \dfrac{{{e}^{4x}}-1}{x} \right)$

This is $\dfrac{0}{0}$ form.

So, by using L-Hospital rule

$\underset{x\to 0}{\mathop{\lim }}\,\left( \dfrac{{{e}^{4x}}\times 4-0}{1} \right)$

$\underset{x\to 0}{\mathop{\lim }}\,\left( 4{{e}^{4x}} \right)$

$=4\times {{e}^{0}}$

$=4\times 1$

$=4$

Hence, this is the answer.

CBSE Questions for Class 11 Commerce Applied Mathematics Limits And Continuity Quiz 5 - MCQExams.com

0:0:1

Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

CBSE Questions for Class 11 Commerce Applied Mathematics Limits And Continuity Quiz 5 - MCQExams.com

0:0:1

Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.